SOLUTION: the sum of the squares of two consecutive even integers is 52. what are the integers? I believe the answers is 4^2 and 6^2 but I can't figure out how to write an equation to show

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Question 148437: the sum of the squares of two consecutive even integers is 52. what are the integers? I believe the answers is 4^2 and 6^2 but I can't figure out how to write an equation to show how I figured it out, Please help!!!
Found 2 solutions by edjones, Earlsdon:
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
Let the smaller be x, the larger x+2
x^2+(x+2)^2=52
x^2+x^2+4x+4=52
2x^2+4x-48=0
2(x^2+2x-24)=0
(x+6)(x-4)=0
x=4
x+2=6
.
Ed

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Try this!
Let x be the first even integer, then (x+2) is the next consecutive even integer.
The sum of their squares is expressed by:
x%5E2%2B%28x%2B2%29%5E2 and this is equal to 52, so you can write the equation:
x%5E2%2B%28x%2B2%29%5E2+=+52 Simply this.
x%5E2%2B%28x%5E2%2B4x%2B4%29+=+52 Combine like-terms and subtract 52 from both sides.
2x%5E2%2B4x-48+=+0 Now you have a quadratic equation that can be solved by factoring after dividing through by 2 to simplify it a bit.
x%5E2%2B2x-24+=+0 Factor.
%28x-4%29%28x%2B6%29+=+0 from which you get:
x+=+4 or x+=+-6
The integers are 4 and 6 or -6 and -4
Check:
4%5E2%2B6%5E2=+16%2B36=52 or...
%28-6%29%5E2%2B%28-4%29%5E2+=+36%2B16=52